3.55 \(\int \sec ^4(a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac {\tan ^5(a+b x)}{5 b}+\frac {\tan ^3(a+b x)}{3 b} \]

[Out]

1/3*tan(b*x+a)^3/b+1/5*tan(b*x+a)^5/b

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2607, 14} \[ \frac {\tan ^5(a+b x)}{5 b}+\frac {\tan ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4*Tan[a + b*x]^2,x]

[Out]

Tan[a + b*x]^3/(3*b) + Tan[a + b*x]^5/(5*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

\begin {align*} \int \sec ^4(a+b x) \tan ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\tan ^3(a+b x)}{3 b}+\frac {\tan ^5(a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 1.81 \[ -\frac {2 \tan (a+b x)}{15 b}+\frac {\tan (a+b x) \sec ^4(a+b x)}{5 b}-\frac {\tan (a+b x) \sec ^2(a+b x)}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4*Tan[a + b*x]^2,x]

[Out]

(-2*Tan[a + b*x])/(15*b) - (Sec[a + b*x]^2*Tan[a + b*x])/(15*b) + (Sec[a + b*x]^4*Tan[a + b*x])/(5*b)

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fricas [A]  time = 0.41, size = 39, normalized size = 1.26 \[ -\frac {{\left (2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right )}{15 \, b \cos \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/15*(2*cos(b*x + a)^4 + cos(b*x + a)^2 - 3)*sin(b*x + a)/(b*cos(b*x + a)^5)

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giac [A]  time = 0.42, size = 26, normalized size = 0.84 \[ \frac {3 \, \tan \left (b x + a\right )^{5} + 5 \, \tan \left (b x + a\right )^{3}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/15*(3*tan(b*x + a)^5 + 5*tan(b*x + a)^3)/b

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maple [A]  time = 0.04, size = 42, normalized size = 1.35 \[ \frac {\frac {\sin ^{3}\left (b x +a \right )}{5 \cos \left (b x +a \right )^{5}}+\frac {2 \left (\sin ^{3}\left (b x +a \right )\right )}{15 \cos \left (b x +a \right )^{3}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6*sin(b*x+a)^2,x)

[Out]

1/b*(1/5*sin(b*x+a)^3/cos(b*x+a)^5+2/15*sin(b*x+a)^3/cos(b*x+a)^3)

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maxima [A]  time = 0.34, size = 26, normalized size = 0.84 \[ \frac {3 \, \tan \left (b x + a\right )^{5} + 5 \, \tan \left (b x + a\right )^{3}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/15*(3*tan(b*x + a)^5 + 5*tan(b*x + a)^3)/b

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mupad [B]  time = 0.39, size = 25, normalized size = 0.81 \[ \frac {{\mathrm {tan}\left (a+b\,x\right )}^3\,\left (3\,{\mathrm {tan}\left (a+b\,x\right )}^2+5\right )}{15\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/cos(a + b*x)^6,x)

[Out]

(tan(a + b*x)^3*(3*tan(a + b*x)^2 + 5))/(15*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**6*sin(b*x+a)**2,x)

[Out]

Timed out

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